
\prob{009C}{分式方程II}

求关于$x$的方程
\[ x^2 + \frac{9x^2}{(x - 3)^2} = 16 \]
的实数根。
\problabels{yellow/代数, green/方程相关问题}

\textit{JYH提供的题目。}

\ans{$x = -1 \pm \sqrt7$}

\subsection{配方}

将等号左侧配方，得
\[ \left(x + \frac{3x}{x - 3}\right)^2 = \frac{6x^2}{x - 3} + 16 \]
左侧括号内通分，得
\[ \left(\frac{x^2}{x - 3}\right)^2 = 6\frac{x^2}{x - 3} + 16 \]
解关于$x^2/(x - 3)$的二次方程，得
\[ \frac{x^2}{x - 3} = 8 \text{或} \frac{x^2}{x - 3} = -2 \]
解两者，得$x = -1 \pm \sqrt7$。
